select t1.user_id id, name, sum_s grade_num
from (
    # 先查询各个用户的积分变化情况并在记录后使用窗口函数添加新列存储最高积分
    select user_id, 
    sum(if(type = 'add', grade_num, (-1)*grade_num)) sum_s,
    max(sum(if(type = 'add', grade_num, (-1)*grade_num)))over() max_sum_s
    from grade_info
    group by user_id
) t1
join user u
on t1.user_id = u.id
where sum_s = max_sum_s     # 只有当前用户的积分变化数值等于积分变化最高值时被保留
order by id;